Question

The activation energy Ea of a reaction is Ea = 43.5Kj.mol. Estimate the variation in the rate k of the reaction when the temperature is raised from 300K to 310K.(R = 8.314 J.K.mol)

Answer 1

The variation in the rate k (k1/k2) is 0.5707 s^-1

Explanation:

Given the following parameters

Activation energy = 43.5Kj.mol

T1 = 300K

T2 = 310K

R = 8.314 J.Kmol

`K\text{ =A}e^{-(Ea)/(RT)}^{}^{}`
`\begin{gathered} \text{where K = rate constant},\text{ Ea = Activation energy, R= gas constant and T = temperature} \\ \ln (k)\text{ = }\ln A\text{ - }(Ea)/(RT) \\ \text{ A is a constant} \\ \text{ if the reaction occurs at two temperature T1 and T2} \\ \ln k1\text{ = }\ln A\text{ - }(Ea)/(RT1)--------\text{ equation 1} \\ \ln k2\text{ = }\ln A\text{ - }(Ea)/(RT2)\text{ --------- EQUATION 2} \\ \text{ substracting equation 1 from 2},\text{ we have} \\ \ln ((k1)/(k2))\text{ = }-(Ea)/(RT1)\text{ + }(Ea)/(RT2) \\ \ln ((k1)/(k2))\text{ = }(Ea)/(R)((1)/(T2)\text{ - }(1)/(T1)) \\ \\ \ln ((k1)/(k2))\text{ = }(43500)/(8.314)((1)/(310)\text{ - }(1)/(300)) \\ \ln ((k1)/(k2))\text{ = 5232.139 }(0.00322\text{ - 0.003333)} \\ \ln ((k1)/(k2))\text{ = 5232.139 (}-\text{ 0.0001071)} \\ \ln ((k1)/(k2))\text{ = -0.560851} \\ \text{Take the exponential of both sides} \\ (k1)/(k2)\text{ = }e^(-0.560851) \\ \frac{k1}{k2\text{ }}=0.5707s^(-1) \end{gathered}`